椭圆x^2/a^2+y^2/b^2=1,F1F2为焦点,P在椭圆上若角F1PF2=60度求e范围_数学解答

椭圆x^2/a^2+y^2/b^2=1,F1F2为焦点,P在椭圆上若角F1PF2=60度求e范围

人气：487 ℃　时间：2019-10-19 14:45:50

解答

当B在短轴顶点时
角F1BF2最大
所以
60<=角F1BF2<180
cos在0到180是减函数
-1F1P=m,F2P=n
m+n=2a
F1F2=2c
cos角F1BF2=(m²+n²-4c²)/2mn
-1<(m²+n²-4c²)/2mn<=1/2
m²+n²+2mn=4a²
m²+n²=4a²-2mn>=2mn
mn<=a²
(m²+n²-4c²)/2mn=(4a²-2mn-4c²)/2mn
=(2a²-2c²)/mn-1
mn<=a²
1/mn>=1/a²
所以(2a²-2c²)/mn-1>=(2a²-2c²)/a²-1=1-2c²/a²
即-1<1-2c²/a²<=1/2
-1/2<=2c²/a²-1<=1
1/4<=c²/a²<=1
1/4<=e²<=1
1/2<=e<=1